Question: $f$ is the Maclaurin series $\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{6n}}{(2n)!}$. $f\left(\sqrt[3]{\dfrac{\pi}{2}}\right)=$
Solution: Note that the terms alternate in sign and have even factorials in the denominator. This suggests that we can use the Maclaurin series for $\cos x$ as a guide. Recall that $\cos x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}\,$. Hence $\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{6n}}{(2n)!}=\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n}}{(2n)!}=\cos \big(x^3\big)$ Now substitute $~x = \sqrt[3]{\frac{\pi}{2}}\,$. $\cos \big(x^3\big)= \cos\left(\sqrt[3]{\frac{\pi}{2}}^3\right)=\cos \frac{\pi}{2}=0$